1
$\begingroup$

We offer two estimators for the average concentration $μ$ of lead in the atmosphere of a region of Quebec where factories manufacturing dyes are located. The first estimator $θ_1$ has a bias equal to $0.2$ and a variance of $0.02$. The second estimator $θ_2$ is unbiased and has a variance equal to $0.06$.

Which one is the best estimator?

I think $θ_2$ is better than $θ_1$ to estimate $μ$, but I am not sure.

EDIT

A PhD student in statistics explained to me that if $MSE(\theta_1) = MSE(\theta_2)$, then we cannot conclude. In other words, $\theta_2$ is not preferred over $\theta_1$ or inversely. I am not sure about that.

$\endgroup$
2
  • 1
    $\begingroup$ I suppose these estimators have been chosen to have equal mean-squared error on purpose? $\endgroup$ Nov 21, 2020 at 16:06
  • $\begingroup$ @preferred_anon Yes, but the purpose is not well oriented because it causes confusion. $\endgroup$
    – David
    Nov 26, 2020 at 1:17

2 Answers 2

3
$\begingroup$

In classical statistics, an estimator is better than another if its MSE is lower.

In this case

$$MSE(\theta_1)=0.02+0.2^2=0.06$$

$$MSE(\theta_2)=\mathbb{V}[\theta_2]=0.06$$

Being

$$MSE(\theta_1)=MSE(\theta_2)$$

$\theta_2$ is preferred as it is unbiased

$\endgroup$
3
  • $\begingroup$ Can you precise you answer? A PhD student in statistics explained to me that if $MSE(\theta_1) = MSE(\theta_2)$, then we cannot conclude. In other words, $\theta_2$ is not preferred over $\theta_1$ or inversely. I am not sure about that. $\endgroup$
    – David
    Nov 22, 2020 at 16:11
  • $\begingroup$ @David: Given the fact that $MSE(\theta_1)=MSE(\theta_2)$ I do prefer the unbiased estimator VS the biased one..You and yuor friend do not? $\endgroup$
    – tommik
    Nov 22, 2020 at 16:18
  • $\begingroup$ In its context, if you could give only one example where $\theta_2$ is preferred to $\theta_1$, then I will accept your answer and give you the bounty $\endgroup$
    – David
    Nov 23, 2020 at 17:54
2
$\begingroup$

In my experience, a situation like this can end up depending on the context, i.e. what you are trying to estimate. Since you are discussing the average concentration of lead in the atmosphere of a particular region, the results of this could be severe, in which case I have been taught taking the unbiased estimator is a better idea here (since your variance is still quite small).

Hope this helps!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .